*Image of solar corona taken at 211Å by the Atmospheric Imaging Assembly of the Solar Dynamics Observatory (SDO).*As the new solar cycle ramps up much attention again will focus on the Sun's corona. Tnanks to new imaging tools, with increased optical as well as temporal resolution, we ought to be getting unparalled images of the outermost atmosphere of the nearest star. Of particlar relevance here will be the Atmospheric Imaging Assembly, built for the

*Solar Dynamics Observatory*by the Lockheed Martin Solar Astrophysics Laboratory.

The device, since installed, has been taking high resolution images of the corona in multiple wavelengths, from the near to the extreme ultra-violet (UV). Shown in the graphic is one of the representative images taken at 211Å, taken last October. This corresponds to the corona's characteristic temperature of 2 x 10^6 K. It is also the wavelength most sensitive to magnetically active regions (ARs) and we see white lines superimposed on those field lines.

Among the interesting questions pursued in the past century was whether the corona was static or not. In a static case its boundary would be more or less fixed, there'd be no expansion even in times of high solar activity.

Of course, a static corona seems (on first blush) to be quite reasonable but that is why it is necessary to test that this is so. The first one to do this was Sydney Chapman. He began by first assuming the condition for hydrostatic equilibrium applied:

dp/
dr = - r {GM

where G is the usual Newtonian gravitational constant, and rho defines

_{s}/ r^{2}}where G is the usual Newtonian gravitational constant, and rho defines

*the plasma density*for the corona, while M_{s }is the mass of the Sun, and r the distance from the solar center:r = n(m

with n the number density for protons

The coronal pressure (P) is given by:

P = 2 n T

Provided both protons and electrons are assumed to have the same temperature.

The

_{p})with n the number density for protons

The coronal pressure (P) is given by:

P = 2 n T

Provided both protons and electrons are assumed to have the same temperature.

The

*thermal conductivity*of the corona is dominated by electron thermal conductivity and takes the form:k = k

for typical coronal conditions the value of k is about 20 times the value of copper at room temperature.

_{o}T^{5/2}for typical coronal conditions the value of k is about 20 times the value of copper at room temperature.

Now, the

*coronal heat flux density*is:

q =
-k Ñ T

A static corona means

A static corona means

*heat inputs cancel heat outputs*so that the divergence:Ñ

Assuming a spherical symmetry for the corona one can write:

**×**q = 0Assuming a spherical symmetry for the corona one can write:

1/r

^{2}[d/dr (r^{2}k_{o}T^{5/2}dT/dr)] = 0Obviously the preceding assumptions mean there must be some distance where the coronal temperature becomes zero.

From the above eqn. one should be able to show:

d(T

where C is a constant.

The integral is:

^{7/2}) = 7/2 (F T_{o }^{5/2})/ 4 p k_{o}d(1/r) = C d(1/r)where C is a constant.

The integral is:

T

_{o }^{7/2}- T_{ }^{7/2}= C[ 1/R_{o }- 1/r]Now,
set the temperature at infinity (T) to zero and obtain:

C =
R

which fixes the total flux at:

F = 2/7 [4 p R

After another step, one finds:

T(r) = T

this gives the temperature T at a distance from the Sun = r . This is based on using a defined value (say T

_{o}T_{o }^{7/2}which fixes the total flux at:

F = 2/7 [4 p R

_{o}k_{o}T_{o}]After another step, one finds:

T(r) = T

_{o}(R_{o}/ r)^{2/7}this gives the temperature T at a distance from the Sun = r . This is based on using a defined value (say T

_{o}= 2 x 10^{6 }K) at a defined distance, say R_{o}= 7 x 10^{8}m. For example, at the Earth’s distance
(r = 1.5 x 10

This seems fine, until one examines the pressure.

Analogous to the temperature formalism, we have, the pressure p(r) at some distance defined by:

Then we are left with the expression for the pressure:

p(¥) = p(R

where ‘k’ denotes a constant composed of all the constant quantities in the previous eqn. (G, M, m

Substituting the given values into the above, one finds p(R

exp[0] = 1

The reason is that the exponential of a very small and negative valued magnitude ® 0

Then:

p(¥) » p(R

But

This led astrophysicists to conclude an unphysical result, and that

^{11}m) one would find: T = 4.3 x 10^{5}KThis seems fine, until one examines the pressure.

Analogous to the temperature formalism, we have, the pressure p(r) at some distance defined by:

p(r) = p(R_{o}) exp
[7/5 GM_{s} m_{p}/ 2 T(R_{o}) R_{o} {( R_{o}
/ r)^{5/7} – 1}]

_{o}/ ¥) that the first term vanishes.Then we are left with the expression for the pressure:

p(¥) = p(R

_{o}) [exp – 7k/5 * 1/ T(R_{o}) R_{o}]where ‘k’ denotes a constant composed of all the constant quantities in the previous eqn. (G, M, m

_{p}etc)Substituting the given values into the above, one finds p(R

_{o}) multiplied by a factorexp[0] = 1

The reason is that the exponential of a very small and negative valued magnitude ® 0

Then:

p(¥) » p(R

_{o})But

*since the pressure of the coronal base would then be***this can’t be***the same as the value at infinity*!This led astrophysicists to conclude an unphysical result, and that

*. If the static model were accurate, the pressure at infinity should be zero, p(¥) = 0, not a small finite pressure that’s effectively equal to the coronal base pressure. This finding led to the further investigations that disclosed a solar “wind” had to flow outwards from the corona.***the static coronal model couldn’t be accurate**Hopefully, with the new imagery devices in place, we can learn much more about the actual dynamics of the corona in the new solar cycle!

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